 #1
 25
 0
Write the expression for the infinitesimal radial inertal force dfr "look at pic"
Consider the nonuniform rigid disc of Figure 1. The density of the material is variable and depends
on the radial position: p = p (r). Its analytical expression is
p=po+p(1)
and where Po and P(1) are known and constants. Note that the density changes only with the radial coordinate:
for a given r changing θ does not affect the density. It should he observed that equation 1 implies that at
the center of the disk (r = O) the density presents its minimum value (p = Po) and for r = R the density
reaches the maximum value (p = po + P1 R)
The radius of the disc is R and its uniform thickness is h. The generalized coordinate is represented by
the angle θ (see Figure 1). The disc can only rotate with respect to G and the translation is not allowed. The
effects of the gravitational forces are not considered (the disc is supposed to rotate in a horizontal plane).
The angular acceleration dθ and the angular velocity d^2θ are supposed to be nonzero quantities. The generic infinitesimal element of disc is identified by the radial coordinate r (note that 0<r< R) and angular
position θ (note that O<θ< 2∏). The infinitesimal element of disc has mass dm and is subjected to the
infinitesimal radial inertial force dFr and the infinitesimal tangential inertial force dFθ
f=ma
f=ma >>> df(r)=dm*a(r)
a(r)= r"r(θ')^2
dm=dr*dθ*h*d(rho)
∴ dfr=(r(θ')^2)(dm)
is this correct based on the question, my English is not so good so I am not sure if i am answering what the problem is asking.
Also on the dm part I understand that infinitesimal is just a tiny piece of the disk so to find the mass i would need a change in radius a change in angle since the high states constant i leave h but since the mass changes i do d(rho)/dr since it changes with the radius but if i put it in my dm equation i would get something like this
dm=dr*dθ*h*d(rho)/dr >>>> dm=dθ*h*d(rho) ????? is this right?
Consider the nonuniform rigid disc of Figure 1. The density of the material is variable and depends
on the radial position: p = p (r). Its analytical expression is
p=po+p(1)
and where Po and P(1) are known and constants. Note that the density changes only with the radial coordinate:
for a given r changing θ does not affect the density. It should he observed that equation 1 implies that at
the center of the disk (r = O) the density presents its minimum value (p = Po) and for r = R the density
reaches the maximum value (p = po + P1 R)
The radius of the disc is R and its uniform thickness is h. The generalized coordinate is represented by
the angle θ (see Figure 1). The disc can only rotate with respect to G and the translation is not allowed. The
effects of the gravitational forces are not considered (the disc is supposed to rotate in a horizontal plane).
The angular acceleration dθ and the angular velocity d^2θ are supposed to be nonzero quantities. The generic infinitesimal element of disc is identified by the radial coordinate r (note that 0<r< R) and angular
position θ (note that O<θ< 2∏). The infinitesimal element of disc has mass dm and is subjected to the
infinitesimal radial inertial force dFr and the infinitesimal tangential inertial force dFθ
Homework Equations
f=ma
The Attempt at a Solution
f=ma >>> df(r)=dm*a(r)
a(r)= r"r(θ')^2
dm=dr*dθ*h*d(rho)
∴ dfr=(r(θ')^2)(dm)
is this correct based on the question, my English is not so good so I am not sure if i am answering what the problem is asking.
Also on the dm part I understand that infinitesimal is just a tiny piece of the disk so to find the mass i would need a change in radius a change in angle since the high states constant i leave h but since the mass changes i do d(rho)/dr since it changes with the radius but if i put it in my dm equation i would get something like this
dm=dr*dθ*h*d(rho)/dr >>>> dm=dθ*h*d(rho) ????? is this right?
Attachments

9.1 KB Views: 303
Last edited: